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Listed below are questions submitted by users of "From Stargazers to Starships" and the answers given to them. This is just a selection--of the many questions that arrive, only a few are listed. The ones included below are either of the sort that keeps coming up again and again, or else the answers make a special point, often going into details which might interest many users.

The reason for my question comes from my fascination with the 2012 end-of-times scenario to which so many people have succumbed. Much of it is hogwash but as with any crackpot theory there exists the miniscule possibility of truth. I think there is more credibility in the theory that earth could have a brown dwarf twin with it's own small planetary system than the theory of a planetoid body having a huge elliptical orbit such that it only comes round once every 3600 yrs.

Thanks and have a great day!

Reply

Dear Steve

I certainly do not believe that the Maya astronomers--a stone age culture with no telescopes--observed astronomical processes which modern instruments fail to detect. Many people have written about it, for years, and some of the correspondence is linked from the end of
http://www.phy6.org/stargaze/Scalend.htm

I expect many more letters after November, when a film "2012" is due to be released--also when "Sky and Telescope" will feature a discussion of 2012. Can't help it.

A large astronomical object with a period of 3600 years seems too close--it would probably affect planetary motions in a detectable way. Anyway, known distant small planets are very small. The idea that the Sun may have a dim distant companion star is old: see
http://en.wikipedia.org/wiki/Nemesis_(star)So far there exists no real evidence

I am professional java software engineer.. yesterday I read about you on internet so decided to talk with you...
these days I am exploring the google map services... last week I created an application which take any location name on earth and provide me the correct latitude and longitude as found on Google Earth
for example

I take the haversine farmula to get the result from latitude and longitude... [continues with a computer code]

Reply

I am not familiar with the haversine formula, and also wonder why the Earth's radius (6378 km at the equator, 6371 at the pole) does not appear anywhere in your code. I also did not recognize the computer code language, though one could probably decode it.

Here is how I would have done it--assuming the Earth to be a sphere of radius R, which is OK for moderate accuracy. I prefer working in spherical coordinates (r, theta, phi) or (r, θ, φ ), with φ the longitude but θ equal to 90–latitude, so southern latitude is negative.
(See also
http://www.phy6.org/stargaze/Slatlong.htm).

Then a unit vector along the radius from the Earth's center to a point has components

(Ux, Uy, Uz) = (sin θ cosφ, sin θ sinφ, cos θ)
For any given point, all three components can be derived from the latitude and longitude.

Let (Vx, Vy, Vz) and (Wx, Wy, Wz) be unit vectors to the two points whose distance you seek. Then the angle A between them has a cosine equal to the scalar product (where * denotes multiplication):

cos A = (Vx*Wx) + (Vy*Wy) +(Vz*Wz)

The distance D between the points is simply RA with A in radians. If a quarter of the circumference of Earth is 10,000 kilometers (original definition of the meter) and A is in degrees, then
D = 10000 (A/90)

If D is much smaller than R (as in your example) you probably get reasonable precision by assuming the Earth is locally flat and using the theorem of Pythagoras in rectangular coordinates. Each degree of latitude then has length 10000/90 kilometers and each degree of longitude length (10000/90) cos L, with L the average longitude of the points.

Because of refraction, the setting Sun looks a bit flattened. The bottom of the Sun may already be below the horizon (in the "extra" visible part of the sky) but you still see it because of refraction, while the top of the Sun, more distant from the horizon, is refracted by a smaller angle. See table in the first of the letters above.

While I'm at it: Does our sun rotate in any particular direction around the
center of the Milky Way galaxy? Say, as viewed from Polaris?
And:
If we were looking down on the Milky Way as a great disk, what angle would the
Sun's pole and planetary disk subtend? Is that the right word?

Reply

I am not familiar with the theory you describe, but am pretty sure it is incorrect, and not only because it would take enormous energy to eject an Earth-size mass (never mind now Jupiter) from the Sun to the Earth's orbit, or to change its orbit to near circular (rather then hit the Sun on the return trip down the other side of the ellipse).

No, the main reason astronomers discount solar origin of the planetary system is the conservation of angular momentum. All but a few percent of the angular momentum of the solar system reside in the planets: it would be hard to see how this would happen if the planets came from the Sun.

About rotation around the galaxy, see
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/980225a.html
The axis of rotation is quite close to the Earth's equatorial plane, and viewed from some distant galaxy behind the pole star, it would have a clockwise component in that plane.
The pole star is not a good reference point, since it rotates with the galaxy.

http://www.phy6.org/stargaze/Stostars.htm Unfortunately, deriving such orbit requires very precise calculations, and often a "mid-course correction" during the flight (or more than one), because at launch it is very hard to achieve the precise velocity needed.

However, we can estimate the velocity of the solar system's rotation around the center (look up "solar apex") by the average Doppler shift of distant stars. From that, assuming Newton's laws hold and that the attraction is contributed by the stars we see (plus the centeral black hole) the average force of gravity can be calculated, As confirmation, Kepler's third law should be valid for objects rotating around the center of our galaxy, or any other.

The trouble is, Kepler's third law is NOT obeyed for objects rotating around the center of our galaxy (or any other). The common interpretation is that galaxies have greater mass than is visible, that they contain "dark matter" as well. (A minority opinion proposes that the cause is departure from Newton's inverse-squares law at great distances). See

Considering the mass of the Earth, the forces generated by rapid rotation are enormous. George Gamov in "Biography of the Earth" promotes the Gerstenkorn theory that the Earth originally rotated much faster, and broke up to create the Moon--I vaguely recall a rotation period like 4 hours (there certainly exists a limit of this order and any faster rotation would break up the Earth). The tides raised by the Moon then would gradually slow down our rotation, as they still do (very slowly). Today most astronomers believe that Earth and Moon evolved separately at the time when the smaller bodies of the solar nebula joined up to form planets.

By all means, read Hal Clement's book "Mission of Gravity."
http://en.wikipedia.org/wiki/Mission_of_Gravity
It is a science fiction story of life on a planet rotating very rapidly and therefore extremely deformed. No material exists which could endure such conditions without breaking up; but it is fun to read. Hal Clement's full name was Harry Clement Stubbs and he taught physics.